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OleMash
4 days ago
7

"A sample of silicon has an average atomic mass of 28.084amu. In the sample, there are three isotopic forms of silicon. About 92

.22% of silicon atoms are 27.9769amu, which have the mass of 28Si; about 4.68% are 28.9764amu, which have the mass of 29Si, and the remaining isotope, 30Si, has a mass of 29.9737amu. Calculate the percent isotopic composition of 30Si."
Chemistry
1 answer:
lorasvet [956]4 days ago
6 0

Answer: The percentage abundance for _{14}^{30}\textrm{Si} isotope is 3.09 %.

Explanation:

The average atomic mass of an element is calculated by taking the sum of the masses of all isotopes weighted by their respective natural fractional abundances.

To compute average atomic mass, the following formula is applied:

\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i   .....(1)

From the information provided:

Let the fractional abundance for _{14}^{28}\textrm{Si} isotope be 'x'

  • For _{14}^{28}\textrm{Si} isotope:

Mass of _{14}^{28}\textrm{Si} isotope = 27.9769 amu

Percentage abundance of _{14}^{28}\textrm{Si} isotope = 92.22 %

Fractional abundance for _{14}^{28}\textrm{Si} isotope = 0.9222

  • For _{14}^{29}\textrm{Si} isotope:

Mass of _{14}^{28}\textrm{Si} isotope = 28.9764 amu

Percentage abundance for _{14}^{28}\textrm{Si} isotope = 4.68%

Fractional abundance of _{14}^{28}\textrm{Si} isotope = 0.0468

  • For _{14}^{30}\textrm{Si} isotope:

Mass of _{14}^{30}\textrm{Si} isotope = 29.9737 amu

Fractional abundance for _{14}^{30}\textrm{Si} isotope = x

  • The average atomic mass of silicon is 28.084 amu

By inserting these values into equation 1, we derive:

28.084=[(27.9769\times 0.9222)+(28.9764\times 0.0468)+(29.9737\times x)]\\\\x=0.0309

To convert this fractional abundance into a percentage, multiply by 100:

\Rightarrow 0.0309\times 100=3.09\%

This shows that the percentage abundance for _{14}^{30}\textrm{Si} isotope is 3.09 %.

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A student puts 0.020 mol of methyl methanoate into an empty and rigid 1.0 L vessel at 450 K. The pressure is measured to be 0.74
KiRa [971]

Explanation:

Initial moles of ethanoic acid = 0.020 mol

At equilibrium, half of the ethanoic acid molecules have reacted.

Thus, moles of ethanoic acid reacted = 0.020 mol * (50% / 100%)

                                                                     = 0.010 mol

Moles of ethanoic acid remaining = 0.020 mol - 0.010 mol = 0.010 mol

The moles of product (CH3COOH)^{2} gas formed are determined as follows:

0.010 mol CH3COOH * (1 mol (CH3COOH)^{2} / 2 mol CH3COOH)

= 0.005 mol (CH3COOH)^{2}

Consequently, the total moles of gas present in the vessel at equilibrium are 0.010 mol CH3COOH and 0.005 mol (CH3COOH)^{2}

Total gas moles at equilibrium = 0.010 mol + 0.005 mol = 0.015 mol

Next, let’s determine the pressure:

0.020 mol of gas has a pressure of 0.74 atm; so under the same conditions, we find the pressure exerted by 0.015 mol of gas:

P1/n1 = P2/n2

P2 = P1*(n2 / n1)

      = 0.74 atm * (0.015 mol / 0.020 mol)

     = 0.555 atm

4 0
8 days ago
A student is given a sample of a blue copper sulfate hydrate. He weighs the sample in a dry covered porcelain crucible and got a
KiRa [971]

Answer:

There are 5.5668 moles of water for every mole of CuSO₄.

Explanation:

The mass of anhydrous CuSO₄ is:

23.403g - 22.652g = 0.751g.

mass of crucible + lid + CuSO₄ - mass of crucible + lid

Given that the molar mass of CuSO₄ is 159.609g/mol, we calculate the moles:

0.751g ×\frac{1mol}{159,609g} = 4.7052x10⁻³ moles CuSO₄

The mass of water in the initial sample is:

23.875g - 0.751g - 22.652g = 0.472g.

mass of crucible + lid + CuSO₄ hydrate - CuSO₄ - mass of crucible + lid

As the molar mass of H₂O is 18.02g/mol, we find the moles:

0.472g ×\frac{1mol}{18,02g} = 2.6193x10⁻² moles H₂O

The mole ratio of H₂O to CuSO₄ is:

2.6193x10⁻² moles H₂O / 4.7052x10⁻³ moles CuSO₄ = 5.5668

This indicates there are 5.5668 moles of water per mole of CuSO₄.

I hope this is helpful!

5 0
8 days ago
Write the balanced molecular and net ionic equation for the reaction that occurs when the contents of the two beakers are added
alisha [964]

1) reacting hydrochloric acid with nickel:

Balanced molecular equation: Ni(s) + 2HCl(aq) → NiCl₂(aq) + H₂(g).

Ionic equation: Ni(s) + 2H⁺(aq) + 2Cl⁻(aq) → Ni²⁺(aq) + 2Cl⁻(aq) + H₂(g).

Net ionic equation: Ni(s) + 2H⁺(aq) → Ni²⁺(aq) + H₂(g).

In this reaction, nickel undergoes oxidation, changing from an oxidation state of 0 to +2, while hydrogen is reduced from +1 to 0 (H₂).

2) reacting sulfuric acid with iron:

Balanced molecular equation: Fe(s) + H₂SO₄(aq) → FeSO₄(aq) + H₂(g).

Ionic equation: Fe(s) + 2H⁺(aq) + SO₄²⁻(aq) → Fe²⁺(aq) + SO₄²⁻(aq) + H₂(g).

Net ionic equation: Fe(s) + 2H⁺(aq) → Fe²⁺(aq) + H₂(g).

In this scenario, iron is oxidized from an oxidation state of 0 to +2, while hydrogen experiences reduction from +1 to 0 (H₂).

3) hydrobromic acid reacting with magnesium:

Balanced molecular equation: Mg(s) + 2HBr(aq) → MgBr₂(aq) + H₂(g).

Ionic equation: Mg(s) + 2H⁺(aq) + 2Br⁻(aq) → Mg²⁺(aq) + 2Br⁻(aq) + H₂(g).

Net ionic equation: Mg(s) + 2H⁺(aq) → Mg²⁺(aq) + H₂(g).

This reaction sees magnesium oxidized from 0 to +2, and hydrogen reduced from +1 to 0 (H₂).

4) acetic acid reacting with zinc:

Balanced molecular equation: Zn(s) + 2CH₃COOH(aq) → (CH₃COO)₂Zn(aq) + H₂(g).

Ionic equation: Zn(s) + 2H⁺(aq) + 2CH₃COO⁻(aq) → Zn²⁺(aq) + 2CH₃COO⁻(aq) + H₂(g).

Net ionic equation: Zn(s) + 2H⁺(aq) → Zn²⁺(aq) + H₂(g).

Here, zinc gets oxidized from 0 to +2 (Zn²⁺), while hydrogen is reduced from +1 to 0 (H₂).

7 0
5 days ago
Addition of an excess of lead (II) nitrate to a 50.0mL solution of magnesium chloride caused a formation of 7.35g of lead (II) c
KiRa [971]

Answer:

[Cl⁻] = 0.016M

Explanation:

To begin, we analyze the reaction:

Pb(NO₃)₂ (aq) + MgCl₂ (aq) → PbCl₂ (s) ↓  +  Mg(NO₃)₂(aq)

This indicates a solubility equilibrium, resulting in the formation of lead(II) chloride precipitate. The salt can dissociate as follows:

           PbCl₂(s)  ⇄  Pb²⁺ (aq)  +  2Cl⁻ (aq)     Kps

Initial        x

React       s

Eq          x - s              s                  2s

Given that this is an equilibrium scenario, the Kps serves as the constant (Solubility product):

Kps = s. (2s)²

Kps = 4s³ = 1.7ₓ10⁻⁵

4s³ = 1.7ₓ10⁻⁵

s =  ∛(1.7ₓ10⁻⁵. 1/4)

s = 0.016 M

3 0
3 days ago
Calculate the molarity of a 10.0% (by mass) aqueous solution of hydrochloric acid.
VMariaS [1037]

The question is incomplete,the complete question:

Determine the molality of a 10.0% (by weight) solution of hydrochloric acid in water:

a) 0.274 m

b) 2.74 m

c) 3.05 m

d) 4.33 m

e) the solution's density is necessary for calculations

Answer:

The molality for a 10.0% (by weight) hydrochloric acid solution is 3.05 mol/kg.

Explanation:

The solution is a 10.0% (by weight) hydrochloric acid mix.

This means there are 10 grams of HCl in 100 grams of the solution.

Amount of HCl = 10 g

Total mass of solution = 100 g

Total mass of solution = Mass of solute + Mass of solvent

Mass of solvent (water) = 100 g - 10 g = 90 g

Calculate moles of HCl = \frac{10 g}{36.5 g/mol}=0.2740 mol

Mass of water converted to kilograms = 0.090 kg

Molality = \frac{0.2740 mol}{0.090 kg}=3.05 mol/kg

<strongTherefore, the molality of a 10.0% (by weight) hydrochloric acid solution is 3.05 mol/kg.

6 0
7 days ago
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